Respuesta :
Answer:
a) 0.019
b) 0.563
c) x = 1.966 hours
Step-by-step explanation:
E(X) = 1
Exponential random variable's probability function is given as
P(X=x) = λ e^(-λ.x)
The cumulative distribution function is given as
P(X ≤ x) = 1 - e^(-λ.x)
a) The time between the arrivals of small aircraft at a county airport that is exponentially distributed.
But the number of planes that land every hour will be obtained using the Poisson distribution formula.
It is the best for discrete systems.
Poisson distribution formula is given as
P(X = x) = (e^-λ)(λˣ)/x!
λ = 1 aircraft per hour.
The probability that more than three aircraft arrive within an hour = P(X > 3)
P(X > 3) = 1 - P(X ≤ 3) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]
P(X > 3) = 1 - 0.98101 = 0.01899 = 0.019 to 3 d.p
b) If 30 separate one-hour intervals are chosen, what Is the probability that no interval contains more than three arrivals
Probability of one 1-hour interval not containing more than 3 arrivals = 1 - P(X > 3)
= 1 - 0.01899 = 0.98101
Probability that thirty 1-hour intervals will not contain more than 3 arrivals = (0.98101)³⁰ = 0.5626 = 0.563 to 3 d.p
c) Determine the length of an interval of time (In hours) such that the probability that no arrivals occur during the interval is 0.14
We can now use the cumulative distribution function for exponential random variable for this
P(X ≤ x) = 1 - e^(-λ.x)
P(X > x) = 1 - P(X ≤ x)
P(X > x) = e^(-λ.x)
λ = 1, x = ?,
0.14 = e⁻ˣ
e⁻ˣ = 0.14
In e⁻ˣ = In 0.14 = -1.966
-x = -1.966
x = 1.966 hours
Hope this Helps!!!
a) Probability that more than three aircraft arrive within an hour is 0.019
b) Probability that no interval contains more than three arrivals is 0.563
c) The length of an interval of time (In hours) such that the probability that no arrivals occur during the interval is 0.14 is : x = 1.966 hours
The Exponential random variable's probability function is given as
[tex]P(X=x) = k e^{(-k.x)}[/tex]
The cumulative distribution function is given as
[tex]P(X \leq x) = 1 - e^{(-k.x)}[/tex]
Also we know E(X) = 1
a) The time between the arrivals of small aircraft at a county airport that is exponentially distributed.
Using the Poisson distribution formula.
[tex]P(X = x) =\frac{ (e^{-k})(k^x)}{x!}[/tex]
λ = 1 aircraft per hour.
The probability that more than three aircraft arrive within an hour:
P(X > 3) = 1 - P(X ≤ 3) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]
P(X > 3) = 1 - 0.98101 = 0.01899 = 0.019 to 3 d.p
b) Probability of one 1-hour interval not containing more than 3 arrivals:
[tex]1 - P(X > 3)= 1 - 0.01899 \\= 0.98101[/tex]
Probability that thirty 1-hour intervals will not contain more than 3 arrivals is: [tex](0.98101)^{30} = 0.5626\\ = 0.563[/tex]
(c).Now using the cumulative distribution function for exponential random variable :
[tex]P(X \leq x) = 1 --e^{-k.x}\\P(X > x) = 1 - P(X \leq x)\\P(X > x) = e^{-k.x}\\k = 1, x = ?,0.14 = e^{-x}\\e^{-x} = 0.14\\In e^{-x} = In 0.14\\ = -1.966\\-x = -1.966\\x = 1.966 hours[/tex]
Learn more:https://brainly.com/question/22883765
